Titration problem?

INTERNATIONAL JUNIOR SCIENCE OLYMPIAD , FIFTH EDITION , GYEONGNAM , SOUTH KOREAINTERNATIONAL JUNIOR SCIENCE OLYMPIAD , FIFTH EDITION , GYEONGNAM , SOUTH KOREA

A weak acid HA is titrated with a standard base ( NaOH solution) and the following titration curve is obtained.

For a weak acid , the dissociation constant , Ka , is defined as:

HA + H2O <=> A^- + H3O^+

Ka= [A-][H3O+]/[HA]

The realtionship between Ka and pH is :

-lgKa=pH - lg{[A-]/[HA]}

and pH = -lg[H3O+]

Using the information provided determine the dissociation constant ,Ka , for the weak acid HA.

a) 10^{-4,2}

b)10^{-4,5}

c)10^{-7,8}

d)10^{-11,5}

2 Answers
Jun 21, 2018

"Ka"= 10^(-4.5)

Explanation:

The "pKa" of a monoprotic weak acid is the same as the "pH" halfway to its equivalence point. Here's how this shortcut works:

Sodium hydroxide "NaOH" neutralizes the weak acid "HA" to produce "NaA" by the equation

"NaOH" + "HA" to "NaA" + "H"_2"O" color(grey)(" balanced.")

One formula unit of sodium hydroxide reacts with one "HA" molecule to produce one formula unit of "NaA". Thus

n("NaA")=n("HA","reacted")=n("NaOH", "added")

Appearantly
n("HA", "reacted")+ n("HA", "remaining") = n("HA", "initial")

Substituting n("HA","reacted") in the second equation with n("NaA") gives

n("HA","remaining") = n("HA", "initial")- n("NaA")

n("NaA") = n("NaOH") = n("HA", "initial") at the equivalence point where V("NaOH", "added") = 22 color(white)(l) "ml" for this question;

Therefore halfway to that point at the titration midpoint at which 11 color(white)(l) "ml" of "NaOH" has been added, n("NaA") = n("NaOH") = color(navy)(1/2)n("HA", "initial")

n("HA","remaining") = n("HA", "initial")- n("NaA")
color(white)(n("HA","remaining")) =color(navy)(1/2)n("HA", "initial")
color(white)(n("HA","remaining") )=n("NaA")

Additionally, given that

"NaA" (aq) color(purple) to "Na"^(+) + "A"^-

"NaA", a salt, completely disassociates as a strong electrolyte,

n("A"^-)=n("NaA")= n("HA").

"Ka" is defined as the equilibrium constant of the reaction

"HA" (aq) rightleftharpoons "H"^+ (aq) + "A"^(-) (aq)

"Ka"="K"_"eq"=(["H"^+] color(red)(cancel(color(black)(["A"^(-)]))))/(color(red)(cancel(color(black)(["HA"]))))=["H"^+]

given that n("A"^-)=n("NaA")= n("HA") as previously stated.

Therefore

"Ka"=["H"^+]=10^(-"pH")=10^(-4.5)

Reference
"Titration Fundamentals", Chemistry LibreText, https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_Fundamentals

See also
"Titration of a Weak Base with a Strong Acid", https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Base_with_a_Strong_Acid

Jun 21, 2018

K_a = 10^(-4.5)

Explanation:

As an alternative approach, you can use the fact that at the half-equivalence point, exactly half of the weak acid has been neutralized by the strong base.

The chemical equation that describes this neutralization reaction looks like this

"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))

At the half-equivalence point, you add enough moles of strong base, which I've represented here by "OH"^(-), to consume half of the initial number of moles of the weak acid.

Since the reaction consumes the weak acid and produces the conjugate base "A"^(-) in a 1:1 mole ratio, you can say that at the half-equivalence point, you have

["HA"] = ["A"^(-)]

Now, notice that at the equivalence point, "22 mL" of the strong base have been added to the weak acid solution. This tells you that it takes "22 mL" of the strong base to consume all the moles of the weak acid present in the initial solution.

So if it takes "22 mL" of the strong base to consume all the moles of the weak acid, it follows that if you add "11 mL" of the strong base, you will consume exactly half of the moles of the weak acid.

This means that the half-equivalence point corresponds to "11 mL" of the strong base added and a "pH" of 4.5.

The question provides you with the equation

-"lg"(K_a) = "pH" - "lg"( (["A"^(-)])/(["HA"]))

which is a version of the Henderson - Hasselbalch equation that can be used to calculate the "pH" of a weak acid - conjugate base buffer.

Plug in the "pH" of the solution at the half-equivalence point and the fact that ["HA"] = ["A"^(-)] to get

- "lg"(K_a)= 4.5 - "lg"(1)

This gets you

"lg"(K_a) = - 4.5

which implies

color(darkgreen)(ul(color(black)(K_a = 10^(-4.5))))

See this answer for a similar question.

SIDE NOTE: The notation used here is based on the fact that

"lg"(x) = log_(10)(x)

See here for more info on that.